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Find the points of intersection of the circles: x^2 + y^2-2x-2y-2=0 and x^2+y^2+2x+2y-2=0. Draw the circles

 Apr 9, 2020
 #1
avatar+1968 
+3
 Apr 9, 2020
 #2
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+1

I don't understand.

Guest Apr 9, 2020
 #3
avatar+1968 
+3

It has a pretty reasonable explanation... Sorry I cannot do this.

 

Maybe this will help more. 

 

https://mathworld.wolfram.com/Circle-CircleIntersection.html

CalTheGreat  Apr 9, 2020
 #4
avatar+111329 
+1

x^2 + y^2-2x-2y-2=0 and x^2+y^2+2x+2y-2=0

 

Setting  these  equal  we  get that

 

x^2 + y^2-2x-2y-2  =   x^2+y^2+2x+2y-2       simplify

 

-2x - 2y   =  2x + 2y

 

-4x  = 4 y

 

-x  =  y

 

Sub  this  into  eithe equation for  y

 

x^2  + (-x)^2  - 2x - 2(-x) - 2  =  0

 

x^2 + x^2  - 2x + 2x  -  2  =  0

 

2x^2  - 2  =  0

 

2x^2  =  2

 

x^2  =  1             take  both roots  and  we have that

 

x = 1       and  x  =  - 1

 

And when x  =1,  y  =-1

 

And when x  = -1, then y = 1

 

So  the points of intersection   are  ( -1, 1)  and (1 , -1)

 

See  the  graphs here :  https://www.desmos.com/calculator/rjkmzcqniv

 

 

cool cool cool

 Apr 9, 2020

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